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\title{悬链线的微分方程}
\author{五六七}
%\date{2025年10月8日}

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\section{导出悬链线的微分方程}
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\begin{frame}[allowframebreaks]{导出悬链线的微分方程}

\vspace{-0.3cm}

一根链子的两段固定在墙上的两个点，其余根据重力自然下垂。求曲线的函数表达式。
用实验得到观测数据，求出函数表达式的参数。

\begin{figure}[ht!]\centering
\includegraphics [height=0.4\textheight, width=0.5\textwidth]{pic/catenary03.png}
\caption{一根真正的悬链线 }
\end{figure}

\begin{figure}[ht!]\centering
\includegraphics [height=0.4\textheight, width=0.6\textwidth]{pic/pantograph-catenary-systemblocks.png}
\caption{悬链线的另一个例子 }
\end{figure}


实验观测：可对实物拍照，然后量出很多点的坐标，然后代入模型函数，求出参数。
或将一张白纸放在墙上，挂一根线，然后描出悬链线的形状，然后同样的数据分析。

%\includegraphics [height=0.5\textheight, width=0.8\textwidth]{pic/pantograph-catenary-systemblocks.png}

%source: \url{https://www.researchgate.net}%/figure/Pantograph-catenary-systemblocks_fig2_220625832


{\color{red}例子2. 导出悬链线满足的微分方程、边界条件、并求解。} 

\begin{center}
\includegraphics [height=0.6\textheight, width=0.8\textwidth]{pic/catenary.png}
\end{center}

设悬链线的方程为 $y(x)$, 设两端的坐标为 $(x_1,y_1)$ 与 $(x_2,y_2)$.

设悬链线上相邻两点的坐标分别为 $P(x,y(x))$ 和 $Q(x+\Delta x, y(x+\Delta x))$. 

设$PQ$之间的这段弧长为 $\Delta s$, 其中 $s$ 是从左端点 $P_1$ 到$P$的弧长。

设单位长度悬链线受到的重力为 $\gamma$, 则弧 $PQ$ 受到的重力为 $W=\gamma \Delta s$. 

设在 $P$ $Q$ 两点的张力分别为 $F_1$ 和 $F_2$, 这是两个向量。

设张力在水平方向上的分量分别为 $H_1=H(x)$ 和 $H_2=H(x+\Delta x)$. 

设张力在竖直方向上的分量分别为 $V_1=V(x)$ 和 $V_2=V(x+\Delta x)$. 

根据水平方向受力平衡，可得 $H_1=H_2$, 记为 $H_0$. 

根据竖直方向受力平衡，可得 $V_1+W=V_2$.  由此将导出悬链线方程。


从 $V_2 - V_1 = W$ 可得 $V(x+\Delta x) - V(x) = \gamma \Delta s$. 
两边除以 $\Delta x$ 可得 $$\frac{V(x+\Delta x) - V(x)}{\Delta x} = \gamma \frac{\Delta s}{\Delta x}. $$

根据勾股定理，可得 $\Delta s = \sqrt{(\Delta x)^2 + [y(x+\Delta x)-y(x)]^2}$. 除以 $\Delta x$ 可得
$$\frac{\Delta s}{\Delta x} = \frac{\sqrt{(\Delta x)^2 + [y(x+\Delta x)-y(x)]^2}}{\Delta x}. $$ 

对上述两式求 $\Delta\to 0$ 时的极限，可得 $$V'(x) = \gamma \sqrt{1+[y'(x)]^2}. $$


因为张力的竖直方向与水平方向的比例，正好就是切线的斜率，所以有 $$y'(x) = \frac{V(x)}{H(x)} = \frac{V(x)}{H_0}.$$

从上述两式可得悬链线满足的微分方程 $$ y'' = \frac{V'(x)}{H_0} = \frac{\gamma}{H_0}\sqrt{1+(y')^2}. $$

记 $\frac{\gamma}{H_0}=a$. 可将方程简写成 $$y'' = a\sqrt{1+(y')^2}. $$



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\section{求解悬链线方程}
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\begin{frame}[allowframebreaks]{求解悬链线方程}

\vspace{-0.3cm}

记 $z=y'$, 所求方程化为 $z' = a\sqrt{1+z^2}$. 

分离变量可得 $\frac{dz}{\sqrt{1+z^2}} = adx$. 两边积分可得 %$$\int \frac{dz}{\sqrt{1+z^2}} = 
$$\text{arcsinh} z= ax +C_1. $$

由此可得 $z$ 是 $x$ 的双曲正弦函数 $z=\text{sinh}(ax+C_1)$. 

积分可得 $y$ 是 $x$ 的双曲余弦函数 $y=\frac{1}{a}\cosh (ax+C_1) + C_2$.

其中双曲正弦函数与双曲余弦函数分别为 
$$\text{sinh}(x) = \frac{e^x - e^{-x}}{2},\,\,\, \text{cosh}(x) = \frac{e^x + e^{-x}}{2}. $$ 



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\section{悬链线方程的边值条件}
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\begin{frame}[allowframebreaks]{悬链线方程的边值条件}

\vspace{-0.3cm}

因为悬链线的两端是固定的，所以有边值条件 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
y(x_1) &=& y_1, \\ 
y(x_2) &=& y_2. 
 \end{array}\right. 
\end{eqnarray*}

将边值条件代入解函数，可得
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\cosh (ax_1 + C_1) + C_2 &=& ay_1, \\
\cosh (ax_2 + C_1) + C_2 &=& ay_2. 
 \end{array}\right. 
\end{eqnarray*}

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\section{确定悬链线方程的参数}
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\begin{frame}[allowframebreaks]{确定悬链线方程的参数 $a$}

\vspace{-0.3cm}

设悬链线的长度为 $L$, 则有 $L>\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

根据弧长公式，可得 $$L=\int_{x_1}^{x_2} \sqrt{1+[y'(x)]^2}dx. $$

根据悬链线方程 $y'' = a\sqrt{1+(y')^2}$, 可得上述积分为 $$L = \frac{1}{a}\int_{x_1}^{x_2} y''(x)dx = \frac{1}{a}[y'(x_2)-y'(x_1)].$$

代入 $y'=z=\text{sinh}(ax+C_1)$ 可得 $$L=\frac{1}{a}[\text{sinh}(ax_2+C_1) - \text{sinh}(ax_1+C_1)]. $$


%综合边值条件和上述等式，可得
%\begin{eqnarray*}
%\left\{\begin{array}{rcl}
%\cosh (ax_2 + C_1) - \cosh (ax_1 + C_1) &=& ay_2 - ay_1, \\ 
%\text{sinh}(ax_2+C_1) - \text{sinh}(ax_1+C_1) &=& aL. 
%\end{array}\right. 
%\end{eqnarray*}

根据双曲正弦的和差化积公式，可得
$$ aL = 2\text{sinh}\left[ \frac{a(x_2-x_1)}{2} \right] \text{cosh} \left[ \frac{a(x_2+x_1)+2C_1)}{2} \right]. $$

将两个边值条件相减，并由双曲余弦的和差化积公式，可得
\begin{eqnarray*}
a(y_2-y_1) &=& \text{cosh}(ax_2+C_1) - \text{cosh}(ax_1+C_1) \\ 
&=& 2\text{sinh}\left[ \frac{a(x_2-x_1)}{2} \right] \text{sinh} \left[ \frac{a(x_2+x_1)+2C_1)}{2} \right].
\end{eqnarray*}

根据 $\text{cosh}^2(x) - \text{sinh}^2(x) = 1$ 可得 
$$\frac{a}{2}\sqrt{L^2 - (y_2-y_1)^2} = \text{sinh}^2 \left[ \frac{a(x_2-x_1)}{2} \right]. $$


%两式相除，并使用和差化积公式，可得 
%$$\text{tanh}(C_1+a(x_1+x_2)/2)) = \frac{\text{sinh}(C_1+a(x_1+x_2)/2)}{\text{cosh}(C_1+a(x_1+x_2)/2)} = \frac{y_2-y_1}{L}. $$ 
%
%取反函数可得 $$C_1+a(x_1+x_2)/2 = \text{arctanh}\left( \frac{y_2-y_1}{L} \right). $$
%
%还是两式相减，可以直接求出 $a$. 



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